Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 303: 14

a. $3$ b. $8$

a. Letting $u=2+tan\frac{t}{2}$, we have $du=\frac{1}{2}sec^2(\frac{t}{2})dt$. At the endpoints $u(0)=2+tan\frac{0}{2}=2$ and $u(-\frac{\pi}{2})=2+tan(-\frac{\pi}{4})=1$. The original integral becomes $\int_1^22udu=u^2|_1^2=2^2-1^2=3$ b. Repeat part (a) with $u(\frac{\pi}{2})=2+tan(\frac{\pi}{4})=3$. The integral becomes $\int_1^32udu=u^2|_1^3=3^2-1^2=8$